# What are the primitives of Lusztig's twisted bialgebra $\mathbf{f}$?

I'm trying to understand the coproduct on Lusztig's $\mathbf{f}$ and, apart from the Chevalley generators, I don't know of any more primitive elements. In the $q=1$ case, the Weyl group acts as a Hopf algebra automorphism and maps the Chevalley generators to elements corresponding to the other simple roots, so those are also primitive; but when $q\neq1$, the braid group action is not a coalgebra automorphism, so image of the Chevalley generators under the braid group don't seem to be primitive. Are there more primitive elements, or do the Chevalley generators give a complete basis?

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Suppose that $x\in \mathbf{f}$ and $r(x)=x\otimes 1+1\otimes x$. Recall $\mathbf{f}$ is graded by $\mathbb{N}I$, WLOG $x$ is homogenous, $x\in \mathbf{f}_\nu$. We know we get primitive elements if $\nu=i$ for some $i\in I$ so let us assume that $\nu$ is not of this form.
Then $\mathbf{f}_\nu$ is generated by products of the form $yz$ with $y\in \mathbf{f}_\lambda$, $z\in \mathbf{f}_\mu$ with both $\lambda$ and $\mu$ nonzero. Now we consider the symmetric bilinear form on $\mathbf{f}$ and compute
$$(x,yz)=(r(x),y\otimes z)=0$$
since $x$ is assumed primitive.
This shows that $x$ is orthogonal to all of $\mathbf{f}_\nu$. But the bilinear form is nondegenerate, so $x=0$. Thus the Chevalley generators give a basis of the primitive elements of $\mathbf{f}$.